package leetcode._08_dynic;

import org.junit.Test;

/**
 * @author pppppp
 * @date 2022/4/7 21:11
 * 给你两个单词 word1 和 word2， 请返回将 word1 转换成 word2 所使用的最少操作数  。
 * 你可以对一个单词进行如下三种操作：
 * 插入一个字符
 * 删除一个字符
 * 替换一个字符
 *  
 * 示例 1：
 * 输入：word1 = "horse", word2 = "ros"
 * 输出：3
 * 解释：
 * horse -> rorse (将 'h' 替换为 'r')
 * rorse -> rose (删除 'r')
 * rose -> ros (删除 'e')
 * <p>
 * 示例 2：
 * 输入：word1 = "intention", word2 = "execution"
 * 输出：5
 * 解释：
 * intention -> inention (删除 't')
 * inention -> enention (将 'i' 替换为 'e')
 * enention -> exention (将 'n' 替换为 'x')
 * exention -> exection (将 'n' 替换为 'c')
 * exection -> execution (插入 'u')
 * <p>
 * 提示：
 * 0 <= word1.length, word2.length <= 500
 * word1 和 word2 由小写英文字母组成
 */
public class _72_编辑距离 {
    @Test
    public void T_0() {

        String[][] nums = {
                {"pneumonoultramicroscopicsilicovolcanoconiosis", "ultramicroscopically"},
                {"mart", "karma"},
                {"ab", "bc"}, {"", "a"},
                {"horse", "ros"},
                {"intention", "execution"}};
        int[] ans = {27, 3, 2, 1, 3, 5};
        for (int i = 0; i < nums.length; i++) {
            System.out.println(minDistance(nums[i][0], nums[i][1]) == ans[i]);
        }
    }

    public int minDistance(String word1, String word2) {
        if (word1.length() == 0 || word2.length() == 0) {
            return Math.max(word1.length(), word2.length());
        }
        int[][] dp = new int[word1.length()][word2.length()];
        char[] ch1 = word1.toCharArray();
        char[] ch2 = word2.toCharArray();
        dp[0][0] = ch1[0] == ch2[0] ? 0 : 1;
        for (int i = 1; i < ch1.length; i++) {
            dp[i][0] = ch1[i] == ch2[0] ? dp[i - 1][0] : dp[i - 1][0] + 1;
        }
        for (int i = 1; i < ch2.length; i++) {
            dp[0][i] = ch1[0] == ch2[i] ? dp[0][i - 1] : dp[0][i - 1] + 1;
        }
        for (int i = 1; i < word1.length(); i++) {
            for (int j = 1; j < word2.length(); j++) {
                if (ch1[i] == ch2[j]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1;
                }
            }
        }
        return dp[word1.length() - 1][word2.length() - 1];
    }


}
